检查当前单词是否接近字符串中的单词的有效方法是什么？

Question

考虑以下示例：

1. 示例 1：

   str1 = "wow...it  looks amazing"
   str2 = "looks amazi"
   

   您看到amazi 接近amazing，str2 输入错误，我想编写一个程序告诉我amazi 接近amazing 然后在str2 我将替换amazi 与 amazing

2. 示例 2：

   str1 = "is looking good"
   str2 = "looks goo"
   

   在这种情况下，更新后的str2 将是"looking good"

3. 示例 3：

   str1 = "you are really looking good"
   str2 = "lok goo"
   

   在这种情况下，str2 将是"good"，因为lok 不接近looking（或者即使程序可以在这种情况下将lok 转换为looking，那么这对于我的问题的解决方案来说很好)

4. 示例 4：

   str1 = "Stu is actually SEVERLY sunburnt....it hurts!!!"
   str2 = "hurts!!"
   

   更新后的str2 将是"hurts!!!"

5. 示例 5：

   str1 = "you guys were absolutely amazing tonight, a..."
   str2 = "ly amazin"
   

   更新的str2 将是"amazing"，"ly" 将被完全删除或替换。

这将是什么算法和代码？

也许我们可以通过按字典顺序查看字符并设置一个 阈值如 0.8 或 80% 所以如果 word2 从 str1 获得 80% 的 word1 连续字符，那么我们将 str2 中的 word2 替换为 str1 的单词？ 请问还有其他有效的python代码解决方案吗？

Answer 1

在这种情况下，您可以使用 Jacard 系数。首先，您需要按空格分隔第一个和第二个字符串。之后，对于 str2 中的每个字符串，对 str1 中的每个字符串取 Jacard 系数，然后替换为能提供最高 Jacard 系数的字符串。

您可以使用sklearn.metrics.jaccard_score。

Answer 2

像这样：

str1 = "wow...it looks amazing"
str2 =  "looks amazi"
str3 = []

# Checking for similar strings in both strings:
for n in str1.split():
    for m in str2.split():
        if m in n:
            str3.append(n)

# If found 2 similar strings:
if len(str3) == 2:
    # If their indexes align:
    if str1.split().index(str3[1]) - str1.split().index(str3[0]) == 1:
        print(' '.join(str3))

elif len(str3) == 1:
    print(str3[0])

输出：

looks amazing

根据 OP 给出的条件更新：

str1 = "good..."
str2 =  "god.."
str3 = []

# Checking for similar strings in both strings:
for n in str1.split():
    for m in str2.split():

        # Calculating matching character in the 2 words:
        c = ''
        for i in m:
            if i in n:
                c+=i
        # If the amount of matching characters is greater or equal to 50% the length of the larger word
        # or the smaller word is in the larger word:
        if len(list(c)) >= len(n)*0.50 or m in n:
            str3.append(n)


# If found 2 similar strings:
if len(str3) == 2:
    # If their indexes align:
    if str1.split().index(str3[1]) - str1.split().index(str3[0]) == 1:
        print(' '.join(str3))

elif len(str3) == 1:
    print(str3[0])

Answer 3

有很多方法可以解决这个问题。这个解决了你所有的例子。我添加了一个最小相似度过滤器来只返回更高质量的匹配。这就是允许在最后一个样本中删除“ly”的原因，因为它并不是所有单词的全部关闭。

Documentation

你可以用pip install python-Levenshtein安装levenshtein

import Levenshtein

def find_match(str1,str2):
    min_similarity = .75
    output = []
    results = [[Levenshtein.jaro_winkler(x,y) for x in str1.split()] for y in str2.split()]
    for x in results:
        if max(x) >= min_similarity:
            output.append(str1.split()[x.index(max(x))])
    return output

您提出的每个样本。

find_match("is looking good", "looks goo")

['looking','good']

find_match("you are really looking good", "lok goo")

['looking','good']

find_match("Stu is actually SEVERLY sunburnt....it hurts!!!", "hurts!!")

['hurts!!!']

find_match("you guys were absolutely amazing tonight, a...", "ly amazin")

['amazing']

Answer 4

我用正则表达式完成了它

def check_regex(str1,str2):
    #New list to store the updated value
    str_new = []
    for i in str2:
        # regular expression for comparing the strings
        x = ['['+i+']','^'+i,i+'$','('+i+')']
        for k in x:
            h=0
            for j in str1:
                #Conditions to make sure the word is close enough to the particular word
                if "".join(re.findall(k,j)) == i or ("".join(re.findall(k,j)) in i and abs(len("".join(re.findall(k,j)))-len(i)) == 1 and len(i)!=2):
                    str_new.append(j)
                    h=1
                    break
            if h==1:
                break
    return str_new
import re
str1 = input().split()
str2 = input().split()
print(" ".join(check_regex(str1,str2)))